hdu 5314 动态树
Happy King
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 821 Accepted Submission(s): 179
Problem Description
There are n cities and n−1 roads in Byteland, and they form a tree. The cities are numbered 1 through n. The population in i-th city is pi.
Soda, the king of Byteland, wants to travel from a city u to another city v along the shortest path. Soda would be happy if the difference between the maximum and minimum population among the cities passed is **no larger than** D. So, your task is to tell Soda how many different pairs (u,v) that can make him happy.
Soda, the king of Byteland, wants to travel from a city u to another city v along the shortest path. Soda would be happy if the difference between the maximum and minimum population among the cities passed is **no larger than** D. So, your task is to tell Soda how many different pairs (u,v) that can make him happy.
Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤500), indicating the number of test cases. For each test case:
The first line contains two integers n and D (1≤n≤100000,1≤D≤109).
The second line contains n integers p1,p2,…,pn (0≤pi≤109).
Each of the following n−1 lines describing roads contains two integers u,v (1≤u,v≤n,u≠v) meaning that there is a road connecting city u and city v.
It is guaranteed that the total number of vertices in the input doesn't exceed 5×105.
The first line contains two integers n and D (1≤n≤100000,1≤D≤109).
The second line contains n integers p1,p2,…,pn (0≤pi≤109).
Each of the following n−1 lines describing roads contains two integers u,v (1≤u,v≤n,u≠v) meaning that there is a road connecting city u and city v.
It is guaranteed that the total number of vertices in the input doesn't exceed 5×105.
Output
For each test case, output the number of different pairs that can make Soda happy.
Sample Input
1
3 3
1 2 3
1 2
2 3
Sample Output
6
/*
hdu 5314 动态树
problem:
给你一个树,求有多少对(u,v)使u->v路径上面的最大值减去最小值不大于limit
solve:
最开始想的是用树链剖分维护最大最小值,结果超时 卒。。
然后看别人说动态树比树链剖分快一点,于是去学习两天动态树,然后用其维护最大最小值 卒。。
感觉没什么思路 - -
后来发现别人维护一个树的size.维持树中的最大最小值的差不大于limit
所以先按照权值排序,用两个指针. 如果 当前值-最左边值(l) > limit,就将l从这个树中除去
然后将当前节点连接到树上面,即判断与其相连的点哪些在树上面(可以用个数组来判断).
大致思路如此,然后就是如何维护size. 首先动态树中的重链是一直变化的,通常我们是用splay来维护重链上面的值,
所以需要outsize来维护哪些不在重链上面的点.
而 重链 和 普通链会在ACCESS的时候发生变化,于是在其过程中维护一下.
参考:http://blog.csdn.net/u013368721/article/details/47086899
hhh-2016-08-21 09:16:05
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson ch[0]
#define rson ch[1]
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
using namespace std;
const int maxn = 300100;
const int INF = 1e9+10;
ll ans ;
struct Node* null;
struct Node
{
Node* ch[2] ;
Node* fa;
int Size ;
int Outsize;
int val,rev;
void newnode(int v)
{
val = v;
Size = 1 ;
Outsize = 0;
fa = ch[0] = ch[1] = null ;
rev = 0;
}
void update_rev()
{
if(this == null)
return ;
swap(ch[0],ch[1]);
rev ^= 1;
}
void push_up () {
if(this == null)
return ;
Size = ch[0]->Size + 1 + ch[1]->Size + ch[0]->Outsize + ch[1]->Outsize;
// cout << ch[0]->Size <<" " <<ch[1] <<Size <<endl;
}
void push_down()
{
if(this == null)
return ;
if(rev)
{
ch[0]->update_rev();
ch[1]->update_rev();
rev = 0;
}
}
void link_child ( Node* to , int d )
{
ch[d] = to;
to->fa = this ;
}
int isroot()
{
return fa == null || this != fa->ch[0] && this != fa->ch[1] ;
}
void down()
{
if ( !isroot () ) fa->down () ;
push_down () ;
}
void Rotate ( int d )
{
Node* f = fa ;
Node* ff = fa->fa ;
f->link_child ( ch[d] , !d ) ;
if ( !f->isroot () )
{
if ( ff->ch[0] == f ) ff->link_child ( this , 0 ) ;
else ff->link_child ( this , 1 ) ;
}
else fa = ff ;
link_child (f,d) ;
f->push_up () ;
}
void splay ()
{
down () ;
while ( !isroot () ) {
if ( fa->isroot () ) {
this == fa->ch[0] ? Rotate ( 1 ) : Rotate ( 0 ) ;
} else {
if ( fa == fa->fa->ch[0] ) {
this == fa->ch[0] ? fa->Rotate ( 1 ) : Rotate ( 0 ) ;
Rotate ( 1 ) ;
} else {
this == fa->ch[1] ? fa->Rotate ( 0 ) : Rotate ( 1 ) ;
Rotate ( 0 ) ;
}
}
}
push_up () ;
}
void access()
{
Node* now = this ;
Node* x = null ;
while ( now != null )
{
// cout <<"now:"<< now->val <<" f:" <<x->val <<endl;
now->splay () ;
now->Outsize += (now->ch[1]->Size+now->ch[1]->Outsize);
now->Outsize -= (x->Size + x->Outsize);
now->link_child ( x , 1 );
now->push_up () ;
x = now ;
now = now->fa ;
}
splay() ;
}
void make_root()
{
access();
update_rev();
}
void cut()
{
access();
ch[0]->fa = null;
ch[0] = null;
push_up();
}
void cut(Node* to)
{
make_root();
to->cut();
}
void link(Node* to)
{
make_root();
to->make_root();
fa = to;
// cout << Size <<" " << Outsize <<endl;
ans += (ll)(to->Size + to->Outsize)*(Size+Outsize);
// cout << ans <<endl;
to->Outsize += (Size + Outsize);
push_up();
}
};
Node memory_pool[maxn];
Node* now;
Node* node[maxn];
struct Edge
{
int to,next;
}edge[maxn << 2];
int head[maxn],tot;
int vis[maxn];
void Clear()
{
now = memory_pool;
now->newnode(-INF);
null = now ++;
null->Size = 0;
tot = 0;
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
}
void add_edge(int u,int v)
{
edge[tot].to = v,edge[tot].next = head[u],head[u] = tot ++;
}
struct Point
{
int id,v;
}po[maxn];
bool cmp(Point a,Point b)
{
return a.v < b.v;
}
int main()
{
int T,n,cas = 1,limit;
int x,a,b;
// freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
Clear();
scanf("%d%d",&n,&limit);
ans = 0;
for(int i = 1; i <= n; i++)
{
scanf("%d",&x);
now->newnode(x);
node[i] = now++;
po[i].v = x,po[i].id = i;
}
sort(po+1,po+1+n,cmp);
for(int i = 1; i < n; i++)
{
scanf("%d%d",&a,&b);
add_edge(a,b);
add_edge(b,a);
}
int l= 1;
for(int i =1;i <= n;i++)
{
while(l <= i && po[i].v - po[l].v > limit)
{
int u = po[l].id;
for(int j = head[u];~j;j = edge[j].next)
{
int v = edge[j].to;
if(!vis[v])
continue;
node[u]->cut(node[v]);
}
vis[u] = 0;
++l;
}
int u = po[i].id;
for(int j = head[u];~j; j = edge[j].next)
{
// cout <<edge[j].to <<endl;
int v = edge[j].to;
if(!vis[v])
continue;
node[u]->link(node[v]);
}
vis[u] = 1;
}
printf("%I64d\n",ans*2);
}
return 0;
}
